How do you solve $x^2-5=0$ graphically and algebraically?

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Answer 1 · 2016-03-20T16:59:54

$x=+-sqrt(5)$ exactly, or approximately $+-2.236$ to 3 decimal places

Given: $" " x^2=5=0$

To solve mathematically we need to have just 1 $x$ and for it to be on one side of the equals sign and everything else on the other side.

To get the x term on its own.

Add $color(blue)(5)$ to both sides of the equation

$" "color(brown)(x^2-5color(blue)(+5)=0color(blue)(+5)$

$x^2+0=5$

$x^2=5$

But $x^2$ is the same as $x xx x$ and we need to have just 1 not 2 of them.

Take the square root of both sides

$sqrt(x^2)=sqrt(5)$

So $x=sqrt(5)" "$ which is 2.236 to 3 decimal places

But both $(-2.236 )xx(-2.236 ) = 5" "$

and $" "(+2.236 )xx(+2.236 )=5$

So $x=+-sqrt(5)$ exactly or approximately $+-2.236$ to 3 decimal places

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Tony B

How do you solve x^2-5=0x^2-5=0 graphically and algebraically?